The moment of inertia of a thin uniform ring of mass I kg about an axis passing through the centre and perpendicular to the plane of the ring is `0.25" kg m"^(2)`. Then the diameter of the ring is

Options

A. 0.25m
B. 0.5m
C. 0.75m
D. 1m

Answer

To solve the problem, we need to find the diameter of a thin uniform ring given its moment of inertia and mass. The moment of inertia \( I \) of a thin ring about an axis passing through its center and perpendicular to its plane is given by the formula:

\[ I = mR^2 \]

where:
- \( I \) is the moment of inertia,
- \( m \) is the mass of the ring,
- \( R \) is the radius of the ring.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Mass of the ring, \( m = 1 \, \text{kg} \)
- Moment of inertia, \( I = 0.25 \, \text{kg m}^2 \)

2. **Use the Moment of Inertia Formula**:
Substitute the known values into the moment of inertia formula:
\[ I = mR^2 \]
This gives us:
\[ 0.25 = 1 \cdot R^2 \]

3. **Solve for \( R^2 \)**:
Rearranging the equation:
\[ R^2 = 0.25 \]

4. **Calculate the Radius \( R \)**:
Taking the square root of both sides:
\[ R = \sqrt{0.25} = 0.5 \, \text{m} \]

5. **Find the Diameter \( D \)**:
The diameter \( D \) of the ring is twice the radius:
\[ D = 2R = 2 \times 0.5 = 1 \, \text{m} \]

### Final Answer:
The diameter of the ring is **1 meter**.

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